How to integrate $\int_{-\infty}^{\infty} xe^{-2\lambda |x|} dx$?

Question

I need to integrate $$\int_{-\infty}^{\infty} xe^{-2\lambda |x|} dx$$ We are given that $\lambda$ is positive and real This is my attempt $$\int_{-\infty}^{0} x e^{2\lambda x} dx+ \int_{0}^{\infty} xe^{-2\lambda x} dx$$ Using u-sub $u = {2\lambda x}$ and integrating by parts $$\frac{1}{2\lambda}\int_{-\infty}^{0} ue^{u} du+ \frac{1}{2\lambda}\int_{0}^{\infty} ue^{-u} du$$ $$= \frac{1}{2\lambda}(-1+1) = 0$$ But it seems wrong to me

Answer 1

This derivation is correct.

You can also just notice that the function under the integral is odd i.e. $$f(-x)=-f(x)$$ for every $x$.

So... of course the definite integral from $-\infty$ to $+\infty$ if it is well defined (i.e. if it converges) will be equal to zero.

See also: Even and Odd functions

It is also important to note that the two improper integrals of $f(x)$ (from minus infinity to zero and from zero to plus infinity) converge. Why is this important? Because there are odd functions for which the integral from $-\infty$ to $+\infty$ is not zero but is undefined. A very simple example is $f(x)=x$ which is such an odd function.

Integral of f(x)=x from minus infinity to plus infinity

Answer 2

The integral has to vanish as the integrand is an odd function. Your approach is correct but because of my first point no calculus was needed here.